Example: electron positron annihilation forbids a single photon production, and searching through this forum (Why can't a single photon produce an electron-positron pair?) and a few books, they always reason with this by using the centre of mass frame of reference where the two incoming bodies $e^{+}$ and $e^{-}$ have equal and opposite momentum, hence due to conservation of energy and momentum the final state (photon) cannot exist since a photon with no momentum is impossible.

However considering a frame of reference - stationary - at the point of collision of an electron and a positron incoming at a 90 degree angle to each other, what forbids a single photon production?

  • 1
    $\begingroup$ Well, it is not only a convenient frame to do things in (even for classical mechanics), plus you separate out the interaction problem from the what-you-see-in-the-lab problem. (And, 0.3c is pretty relativistic). $\endgroup$ – Jon Custer Mar 21 at 17:40
  • $\begingroup$ I am not sure about this, the low end of mildly relativistic cases is b>0.3. For b<0.3 the γ is generally approximated as ~1. $\endgroup$ – Gav Chatz Mar 25 at 11:22

You can use any frame you want. If something is forbidden, it is forbdden in all frames. The math is generally easiest in the center-of-mass frame... sometimes so easy that the impossibility of a reaction is obvious. The reason why is that by choosing this frame you have already satisfied one of the conservation laws.

  • $\begingroup$ Thank you for the reply, however in the case of a stationary reference frame at the collision point i dont see something forbidding a single photon generation from the annihilation. The basis of this rule (no single photon) is supported in literature for just the COM reference frame. So how is " If something is forbidden, it is forbidden in all frames" supported in that case? is there something i am missing ? $\endgroup$ – Gav Chatz Mar 25 at 11:30
  • $\begingroup$ I’m sorry, but I can’t figure out what is confusing you. Let me try again. If one photon is produced, then observers in all reference frames see one photon. This includes the COM frame, where the obsever would see that single photon at rest. A photon at rest is impossible, so this scenario cannot happen. $\endgroup$ – G. Smith Mar 25 at 15:46
  • $\begingroup$ Then you could say that when physicists look at any given case they firstly investigate from the COM frame to establish some rules. So it is not that it is just mathematically convenient, nor that i could use any frame i would want. Is that correct? and if yes, where would i look to establish this from literature. not just examples but the piece of theory that proves the COM frame of reference as the "rule-maker". The question remains the same as in the title of the post. $\endgroup$ – Gav Chatz Mar 26 at 17:18
  • $\begingroup$ I think you are looking at this completely wrong. No frame is the “rule-maker”. You can pick any frame you want. Physicists tend to pick the COM frame for mathematical convenience, because conservation of momentum is satisfied there. I’m going to stop further explanation because I am simply saying the same thing over and over. $\endgroup$ – G. Smith Mar 26 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.